Find the length of the cardioid r 1 sin theta

This is the equation of a cardioid. The polar equation is expressed in terms of a periodic function, so we have to know its period to get the exact length (and not just a portion or some excess of it).

1

+

sin

θ

has period

T

=

2

π

(the function can be obtained as a vertical translation of the sine function in the plane of coordinates

(

θ

,

r

)

).

The length of a periodic polar curve can be computed by integrating the arc length on a complete period of the function, i.e. on an interval

I

of length

T

=

2

π

:

l

=

∫

I

d

s

where

d

s

=

√

r

2

+

(

d

r

d

θ

)

2

d

θ

So we have to compute the derivative:

d

r

d

θ

=

d

d

θ

(

1

+

sin

θ

)

=

cos

θ

and this implies

d

s

=

√

(

1

+

sin

θ

)

2

+

(

cos

θ

)

2

d

θ

=

√

1

+

2

sin

θ

+

sin

2

θ

+

cos

2

θ

d

θ

By the Pythagorean trigonometric identity

cos

2

θ

+

sin

2

θ

=

1

:

d

s

=

√

2

+

2

sin

θ

d

θ

=

√

2

(

1

+

sin

θ

)

d

θ

Integrating that square root is not easy... we'd better take advantage of trigonometric identities. We have

√

1

+

sin

θ

(multiplied by a numerical coefficient) and we would like to get rid of that square root. The half angle formula for cosine states that

cos

2

(

α

2

)

=

1

+

cos

α

2

. We can use this with sine too, because if

α

=

θ

−

π

2

then

cos

α

=

cos

(

θ

−

π

2

)

=

sin

θ

. So we substitute in the formula and get

cos

2

(

θ

−

π

2

2

)

=

1

+

cos

(

θ

−

π

2

)

2

and then

2

cos

2

(

θ

2

−

π

4

)

=

1

+

sin

θ

By substitution in the last

d

s

expression that we wrote

d

s

=

√

4

cos

2

(

θ

2

−

π

4

)

d

θ

=

2

∣

∣

∣

cos

(

θ

2

−

π

4

)

∣

∣

∣

d

θ

Now we can integrate this on an interval of length

2

π

. To do that, we would like to get rid of the absolute value. The graph of the integrand function is the following

graph{2*abs(cos(x/2-pi/4)) [-10, 10, -5, 5]}

We know that

cos

(

x

2

−

π

4

)

≥

0

⇔

−

π

2

+

2

k

π

≤

x

2

−

π

4

≤

π

2

+

2

k

π

for

k

any integer value, so we get that

−

π

2

π

+

4

k

π

≤

x

≤

3

2

π

+

4

k

π

, and in particular

−

π

2

π

≤

x

≤

3

2

π

.

In the end, we can integrate the arc length on

I

=

[

−

π

2

,

3

2

π

]

, since this is an interval of length

2

π

. We can remove the absolute value, since

cos

(

x

2

−

π

4

)

≥

0

on

I

.

∫

3

2

π

−

π

2

2

cos

(

θ

2

−

π

4

)

d

θ

=

4

∫

3

2

π

−

π

2

1

2

cos

(

θ

2

−

π

4

)

d

θ

=

4

sin

(

θ

2

−

π

4

)

∣

3

2

π

−

π

2

=

4

sin

(

π

2

)

−

4

sin

(

−

π

2

)

=

4

+

4

=

8